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Question
The value of x which satisfies ∣∣
∣∣x−22x−33x−4x−42x−93x−16x−82x−273x−64∣∣
∣∣=0 is:
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Solution
Δ=∣∣
∣∣x−22x−33x−4x−42x−93x−16x−82x−273x−64∣∣
∣∣=0
Applying R2→R2−R1 & R3→R3−R1, then we get
Δ=∣∣
∣∣x−22x−33x−4−2−6−12−6−24−60∣∣
∣∣=0
⇒Δ=∣∣
∣∣x−22x−33x−41361410∣∣
∣∣=0
Expanding along 1st row, we have:
(x−2).6−(2x−3).4+(3x−4).1=0
⇒x=4
Applying R2→R2−R1 & R3→R3−R1, then we get
Δ=∣∣ ∣∣x−22x−33x−4−2−6−12−6−24−60∣∣ ∣∣=0
⇒Δ=∣∣ ∣∣x−22x−33x−41361410∣∣ ∣∣=0
Expanding along 1st row, we have:
(x−2).6−(2x−3).4+(3x−4).1=0
⇒x=4
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